Solve for $x$ and $y$ using substitution. ${2x-y = 7}$ ${x = 6y-2}$
Explanation: Since $x$ has already been solved for, substitute $6y-2$ for $x$ in the first equation. ${2}{(6y-2)}{- y = 7}$ Simplify and solve for $y$ $12y-4 - y = 7$ $11y-4 = 7$ $11y-4{+4} = 7{+4}$ $11y = 11$ $\dfrac{11y}{{11}} = \dfrac{11}{{11}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = 6y-2}\thinspace$ to find $x$ ${x = 6}{(1)}{ - 2}$ $x = 6 - 2$ ${x = 4}$ You can also plug ${y = 1}$ into $\thinspace {2x-y = 7}\thinspace$ and get the same answer for $x$ : ${2x - }{(1)}{= 7}$ ${x = 4}$